To elaborate on the earlier response:

`(log_10 x)^2 = log_10 4*log_10 6` has two solutions.

`log_10 x = +-sqrt(log_10 4*log_10 6)`

The negative value gives another solution of the equation.

`log_10 x = -sqrt(log_10 4*log_10 6)`

=> `x = 10^(-sqrt(log_10 4*log_10 6))`

=> `x ~~ 0.20679`

The equation `log_4 x = log_x 6` has to be solved for x.

Use the formula `log_a b = (log_x b)/(log_x a)` which can be used to change the base of any logarithm.

`log_4 x = log_x 6`

Change both the logarithm on both the sides to a common base, for example 10

=> `(log_10 x)/(log_10 4) = (log_10 6)/(log_10 x)`

=> `(log_10 x)^2 = log_10 4*log_10 6`

=> `log_10 x = sqrt(log_10 4*log_10 6)`

=> `x = 10^sqrt(log_10 4*log_10 6)`

=> `x ~~ 4.835`

**The solution of the equation is `x = 10^sqrt(log_10 4*log_10 6) ~~ 4.835`**